{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Physics 6th Edition Solutions Serway Chapter 16 Solutions

Physics 6th Edition Solutions Serway Chapter 16 Solutions -...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 16 Electrical Energy and Capacitance Quick Quizzes 1. (b). The field exerts a force on the electron, causing it to accelerate in the direction opposite to that of the field. In this process, electrical potential energy is converted into kinetic energy of the electron. Note that the electron moves to a region of higher potential, but because the electron has negative charge this corresponds to a decrease in the potential energy of the electron. 2. (b), (d). Charged particles always tend to move toward positions of lower potential energy. The electrical potential energy of a charged particle is PE qV = and, for positively- charged particles, this increases as V increases. For a negatively-charged particle, the potential energy decreases as V increases. Thus, a positively-charged particle located at would move toward the left. A negatively-charged particle would oscillate around which is a position of minimum potential energy for negative charges. xA = xB = 3. (d). If the potential is zero at a point located a finite distance from charges, negative charges must be present in the region to make negative contributions to the potential and cancel positive contributions made by positive charges in the region. 4. (c). Both the electric potential and the magnitude of the electric field decrease as the distance from the charged particle increases. However, the electric flux through the balloon does not change because it is proportional to the total charge enclosed by the balloon, which does not change as the balloon increases in size. 5. (a). From the conservation of energy, the final kinetic energy of either particle will be given by () ( ) 0 fi if i f f i KE KE PE PE qV qV q V V q V =+ − = +−= − −= For the electron, qe giving and 1 V V =− =+ ( )( ) 1 V 1 eV f KE e =−− + . For the proton, , so and 1 V V ( )( ) 1 V 1 eV e f KE = −− = + , the same as that of the electron. 6. (c). The battery moves negative charge from one plate and puts it on the other. The first plate is left with excess positive charge whose magnitude equals that of the negative charge moved to the other plate. 35
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
36 CHAPTER 16 7. (a) C decreases. (b) Q stays the same. (c) E stays the same. (d) V increases. (e) The energy stored increases. Because the capacitor is removed from the battery, charges on the plates have nowhere to go. Thus, the charge on the capacitor plates remains the same as the plates are pulled apart. Because 00 QA E σ == ∈∈ , the electric field is constant as the plates are separated. Because V = Ed and E does not change, V increases as d increases. Because the same charge is stored at a higher potential difference, the capacitance has decreased. Because 2 2 QC = Energy stored and Q stays the same while C decreases, the energy stored increases. The extra energy must have been transferred from somewhere, so work was done. This is consistent with the fact that the plates attract one another, and work must be done to pull them apart.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 34

Physics 6th Edition Solutions Serway Chapter 16 Solutions -...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon bookmark
Ask a homework question - tutors are online